Explanation Given information: The effective length of the column is l e = 22 ft . The adjusted allowable stress for compression is σ C = 1050 psi . The adjusted modulus of elasticity is E = 440 × 10 3 psi . The centric load acting in the column is P = 18 kip . The width of the column is b = 2 d . Calculation: Find the cross sectional area ( A ) of the column using the equation. A = b d Here, the width of the column is b and the depth of the column is d . Substitute 2 d for b . A = ( 2 d ) d = 2 d 2 Find the radio of the effective length to the dimension of the cross section ( L e d ) . L e d = 22 ft × 12 in . 1 ft d = 264 d Find the stress ( σ C E ) using the equation. σ C E = 0.822 E ( L e d ) 2 Substitute 440 × 10 3 psi for E and 264 d for L e d . σ C E = 0.822 × 440 × 10 3 ( 264 d ) 2 = 361 , 680 d 2 69 , 696 Find the column stability factor C p using the Equation. C p = 1 + ( σ C E / σ C ) 2 c − [ 1 + ( σ C E / σ C ) 2 c ] 2 − σ C E / σ C c Here, the allowable stress for compression grain is σ C . For sawn lumber column, the value of c is c = 0.8 . Substitute 361 , 680 d 2 69 , 696 for σ C E , 1,050 psi for σ C , and 0.8 for c . C p = 1 + ( 361 , 680 d 2 69 , 696 1 , 050 ) 2 × 0.8 − [ 1 + ( 361 , 680 d 2 69 , 696 1 , 050 ) 2 × 0.8 ] 2 − 361 , 680 d 2 69 , 696 1 , 050 0.8 = 1 + ( 361 , 680 d 2 73 , 180 , 800 ) 2 × 0.8 − [ 1 + ( 361 , 680 d 2 73 , 180 , 800 ) 2 × 0.8 ] 2 − 361 , 680 d 2 58 , 544 , 640 Calculate the allowable stress ( σ a l l ) using the relation. σ a l l = σ C C P Substitute 1,050 psi for σ C and 1 + ( 361 , 680 d 2 73 , 180 , 800 ) 2 × 0

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ISBN: 9780073398235

Author: Ferdinand P. Beer, E. Russell Johnston Jr., John T. DeWolf, David F. Mazurek

Publisher: McGraw-Hill Education

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Chapter 10 Solutions

Mechanics of Materials, 7th Edition

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Solution Summary: The author calculates the cross sectional area ( A ) of the column using the equation.

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